class matrix:

    def __init__(self, a):
        self.a = a

    def __repr__(self):
        return f"{self.a[0][0]} {self.a[1][0]}"

def unit():
    return matrix([[1, 0], [0, 1]])

def mul(A, B):
    C = matrix([[0, 0], [0, 0]])
    for i in range(2):
        for j in range(2):
            for k in range(2):
                C.a[i][j] += A.a[i][k] * B.a[k][j]
    return C

def pow(A, n):
    res, tmp = unit(), A
    while n:
        # 一定要注意，矩阵乘法是不满足乘法交换律的！
        if n & 1: res = mul(res, tmp)
        tmp = mul(tmp, tmp)
        n >>= 1
    return res

a, b = map(int, input().split())
x, y = map(int, input().split())
k = int(input())
x, y = x / 100, y / 100
res, A = matrix([[1 - x, y], [x, 1 - y]]), matrix([[a, 0], [b, 0]])
res = mul(pow(res, k), A)
print(res)
'''
from random import randint as r
aa = [(100, 200), (4000, 10000), (int(1e6), int(1e7)), (100, 200)]
bb = [(100, 200), (4000, 10000), (int(1e6), int(1e7)), (int(9e7), int(1e8))]
xy = (0, 100)
kk = [(100, 200), (40000, 100000), (int(1e6), int(1e7)), (int(9e8), int(1e9))]
for ww in range(len(aa)):
    a, b = r(*aa[ww]), r(*bb[ww])
    x, y = r(*xy), r(*xy)
    k = r(*kk[ww])
    s_i = f'{a} {b}\n{x} {y}\n{k}'
    print(s_i)
    with open(f"{ww + 1}.in", 'w') as f:
        f.write(s_i)
'''
'''
| a |    | a - a * x + b * y |
| b | -> | b - b * y + a * x |

可以由这个来乘
| 1-x  y  |
|  x  1-y |

          | 1-x  y  |      | a 0 |
所以res = |  x  1-y |, A = | b 0 |,

                k
       __________________
      |                  |
res = res * res * ... * res * A就可以统计k次倒水后的容量了

好像这次看递推矩阵的时候理解的快一些，好一些了？以及这题好像要开spj，好麻烦的样子。。
'''
# -------------------------------------------------------------------------------

# 比赛环境为python3.8.6版本自带的IDLE，最好早点熟悉一下。。这个东西的提示时有时无

# 菜单栏最右边的Help的Python Docs 比赛时候也可以看，不过建议还是提前多了解了解，

# 比赛的时候至少知道在文档找什么能用的上。

